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Use the given information to answer the following questions.

center (2, -11, 6), radius 10

(a) Find an equation of the sphere with the given center and radius.


(b) What is the intersection of this sphere with the yz-plane?
, x = 0


(c) What is the intersection of this sphere with the xy-plane?
, z = 0

User Abbi
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1 Answer

2 votes

Answer:

a)100

b) (0, -11 + √(84 - (z - 6)^2), z) and (0, -11 - √(84 - (z - 6)^2), z)

c)(2 + √(64 - (y + 11)^2), y, 0) and (2 - √(64 - (y + 11)^2), y, 0)

Explanation:

a)

An equation of a sphere with center (a, b, c) and radius r has the form:

(x- a)^2 + (y - b)^2 + (z - c)^2 = r^2

We are given the center of the sphere as (2, -11, 6) and the radius as 10. Substituting these values into the equation, we get:

(x - 2)^2 + (y + 11)^2 + (z - 6)^2 = 100

So the equation of the sphere is (x - 2)^2 + (y + 11)^2 + (z - 6)^2 = 100.

b) To find the intersection of the sphere with the yz-plane (x = 0), we substitute x = 0 into the equation of the sphere:

(0 - 2)^2 + (y + 11)^2 + (z - 6)^2 = 100

(y + 11)^2 + (z - 6)^2 = 84

This is the equation of a circle with center (0, -11, 6) and radius √84. To find the point(s) of intersection with the yz-plane, we set x = 0 in the equation of the sphere and solve for y and z:

(y + 11)^2 + (z - 6)^2 = 84

y = -11 ± √(84 - (z - 6)^2)

So the two points of intersection with the yz-plane are

(0, -11 + √(84 - (z - 6)^2), z) and (0, -11 - √(84 - (z - 6)^2), z).

c)

To find the intersection of the sphere with the xy-plane (z = 0), we substitute z = 0 into the equation of the sphere:

(x - 2)^2 + (y + 11)^2 + (0 - 6)^2 = 100

(x - 2)^2 + (y + 11)^2 = 64

This is the equation of a circle with center (2, -11) and radius 8. To find the point(s) of intersection with the xy-plane, we set z = 0 in the equation of the sphere and solve for x and y:

(x - 2)^2 + (y + 11)^2 = 64

x = 2 ± √(64 - (y + 11)^2)So the two points of intersection with the xy-plane are (2 + √(64 - (y + 11)^2), y, 0) and (2 - √(64 - (y + 11)^2), y, 0).

User Race
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